# Eigenvalues of a Hermitian Matrix are Real Numbers

## Problem 202

Show that eigenvalues of a Hermitian matrix $A$ are real numbers.

(*The Ohio State University Linear Algebra Exam Problem*)

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We give two proofs. These two proofs are essentially the same.

The second proof is a bit simpler and concise compared to the first one.

Contents

## Proof 1.

Let $\lambda$ be an arbitrary eigenvalue of a Hermitian matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.

Then we have

\[ A\mathbf{x}= \lambda \mathbf{x}. \tag{*}\]
Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we obtain

\begin{align*}

\bar{\mathbf{x}}^{\trans}(A\mathbf{x}) &= \bar{\mathbf{x}}^{\trans}(\lambda \mathbf{x})\\

&=\lambda \bar{\mathbf{x}}^{\trans}\mathbf{x}\\&

=\lambda ||\mathbf{x}||.

\end{align*}

We also have

\begin{align*}

\bar{\mathbf{x}}^{\trans}(A\mathbf{x})=(A\mathbf{x})^{\trans} \bar{\mathbf{x}}=\mathbf{x}^{\trans}A^{\trans}\bar{\mathbf{x}}.

\end{align*}

The first equality follows because the dot product $\mathbf{u}\cdot \mathbf{v}$ of vectors $\mathbf{u}, \mathbf{v}$ is commutative.

That is, we have

\[\mathbf{u}\cdot \mathbf{v}=\mathbf{u}^{\trans}\mathbf{v}=\mathbf{v}^{\trans}\mathbf{u}=\mathbf{v}\cdot\mathbf{u}.\]
We applied this fact with $\mathbf{u}=\bar{\mathbf{x}}$ and $\mathbf{v}=A\mathbf{x}$.

Thus we obtain

\[\mathbf{x}^{\trans}A^{\trans}\bar{\mathbf{x}}=\lambda ||\mathbf{x}||.\]
Taking the complex conjugate of this equality, we have

\[\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}=\bar{\lambda} ||\mathbf{x}||. \tag{**}\]
(Note that $\bar{\bar{\mathbf{x}}}=\mathbf{x}$. Also $\overline{||\mathbf{x}||}=||\mathbf{x}||$ because $||\mathbf{x}||$ is a real number.)

Since the matrix $A$ is Hermitian, we have $\bar{A}^{\trans}=A$.

This yields that

\begin{align*}

\bar{\lambda} ||\mathbf{x}|| &\stackrel{(**)}{=} \bar{\mathbf{x}}^{\trans}A\mathbf{x}\\

& \stackrel{(*)}{=}\bar{\mathbf{x}}^{\trans} \lambda \mathbf{x}\\

&=\lambda ||\mathbf{x}||.

\end{align*}

Recall that $\mathbf{x}$ is an eigenvector, hence $\mathbf{x}$ is not the zero vector and the length $||\mathbf{x}||\neq 0$.

Therefore, we divide by the length $||\mathbf{x}||$ and get

\[\lambda=\bar{\lambda}.\]

It follows from this that the eigenvalue $\lambda$ is a real number.

Since $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that all the eigenvalues of the Hermitian matrix $A$ are real numbers.

## Proof 2.

Let $\lambda$ be an arbitrary eigenvalue of a Hermitian matrix $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$.

Then we have

\[ A\mathbf{x}= \lambda \mathbf{x}. \tag{*}\]

Multiplying by $\bar{\mathbf{x}}^{\trans}$ from the left, we obtain

\begin{align*}

\bar{\mathbf{x}}^{\trans}(A\mathbf{x}) &= \bar{\mathbf{x}}^{\trans}(\lambda \mathbf{x})\\

&=\lambda \bar{\mathbf{x}}^{\trans}\mathbf{x}\\&

=\lambda ||\mathbf{x}||.

\end{align*}

Now we take the conjugate transpose of both sides and get

\[\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}=\bar{\lambda}||\mathbf{x}||. \tag{***}\]
Since $A$ is a Hermitian matrix, we have $\bar{A}^{\trans}=A$.

Then the left hand side becomes

\begin{align*}

\bar{\mathbf{x}}^{\trans}\bar{A}^{\trans}\mathbf{x}&=\bar{\mathbf{x}}^{\trans}A\mathbf{x}\\

& \stackrel{(*)}{=} \bar{\mathbf{x}}^{\trans}\lambda\mathbf{x}\\

&=\lambda ||\mathbf{x}||. \tag{****}

\end{align*}

Therefore comparing (***) and (****) we obtain

\[\lambda ||\mathbf{x}||=\bar{\lambda}||\mathbf{x}||.\]
Since $\mathbf{x}$ is an eigenvector, it is not the zero vector and the length $||\mathbf{x}||\neq 0$.

Dividing by the length $||\mathbf{x}||$, we obtain $\lambda=\bar{\lambda}$ and this implies that $\lambda$ is a real number.

Since $\lambda$ is an arbitrary eigenvalue of $A$, we conclude that every eigenvalue of the Hermitian matrix $A$ is a real number.

## Corollary

Every real symmetric matrix is Hermitian. Thus, as a corollary of the problem we obtain the following fact:

## Related Question.

**Problem**. Let $A$ be an $n\times n$ real symmetric matrix.

Prove that there exists an eigenvalue $\lambda$ of $A$ such that for any vector $\mathbf{v}\in \R^n$, we have the inequality

\[\mathbf{v}\cdot A\mathbf{v} \leq \lambda \|\mathbf{v}\|^2.\]

Note that the inequality makes sense because eigenvalues of $A$ are real by Corollary.

For a proof of this problem, see the post “Inequality about Eigenvalue of a Real Symmetric Matrix“.

## The Ohio State University Linear Algebra Exam Problems and Solutions

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